a3+a8=a4+a7=13a4*a7=22,由上2式及递增等差数列a4=2,a7=11,公差=（11-2）/（7-4）=3所以首项a1=2-3*3=-7an=3n-10sn=(-7)*2^1+(-4)*2^2+(-1)*2^3+……+(3n-10)*2^n2sn= (-7)*2^2+(-4)*2^3+……+(3n-13)*2^n+(3n-10)*2...第二问呢。。。。sn=(-7)*2^1+(-4)*2^2+(-1)*2^3+……+(3n-10)*2^n2sn=(-7)*2^2+(-4)*2^3+……+(3n-13)*2^n+(3n-10)*2^(n+1)相减得，(3n-10)*2*(n+1)+14-(3*2^2+3*2^3+……)=(6n-26)*2^n+26=(3n-13)*2^(n+1)+26 不知是否有算错。。